Past exam of the mathematics course of the University of Cambridge / 2019 / ia / Paper 2 / 5C / c

The identityshows that $y_n^{(p)}=h^3{n}^3/6$ is a particular solution. Combining it with the homogeneous solution from part (b) givesThe condition $y_0=0$ gives $A=0$, and $y_{-1}=y_1$ gives $B=-h^3/6$. HenceThe exact initial-value solution is $y(t)=t^3/6$. Since $N=1/h$,Thus the absolute endpoint error is $h^2/6$, consistent with the second-order truncation error of the centred scheme.